Let $X$ be a spectrum and $H_p(X;\Omega_q^{Spin})\Rightarrow MSpin_{p+q}(X)$ the Atiyah-Hirzebruch spectral:

The differential $d_2\colon H_p(X;\Omega_1^{Spin})\to H_{p-2}(X;\Omega_2^{Spin})$ is the dual of $Sq^2\colon H^{p-2}(X;\mathbb{Z}_2)\to H^p(X;\mathbb{Z}_2)$

The differential $d_2\colon H_p(X;\Omega_0^{Spin})\to H_{p-2}(X;\Omega_1^{Spin})$ is reduction mod $2$ (denoted with $r$) composed with the dual of $Sq^2$

And the proof is based on the following observation: $d_2$ is a stable homology operation and thus induced from elements in $$[H\mathbb{Z}/2, \Sigma^2 H\mathbb{Z}/2]\cong \mathbb{Z}/2\langle Sq^2\rangle$$ or $$[H\mathbb{Z}, \Sigma^2 H\mathbb{Z}/2]\cong \mathbb{Z}/2\langle Sq^2\circ r\rangle$$

where $HR$ is the Spectrum representing singular cohomology with coefficient $R$.

In this paper, def 3.1 suggest that an Homology Operation is a natural transformation between homology functors, but for example Switzer introduces the Homology Cooperations in a complete different way (they might be different but it seems that in some cases they re the dual of the cohomology operations). So my questions are:

1) First of all, what's the relationship between homology operations and the homology cooperations described in Switzer's book for example.

2) Why are we looking at the stable Cohomology operations and then dualise it?

3) later in the paper the following reasoning is done: the edge homomorphism for the AHSS for stable homotopy is a stable homology operation form stable homotopy to singular homology, i.e. an element of $[S^0,H\mathbb{Z}]\cong \mathbb{Z}\langle h \rangle$ where $h$ is the Hurewicz homomorphism. Therefore (after testing with certain space) we can conclude that the edge homomorphism is given by the Hurewicz map. Where can I see some reference for the fact that $[S^0,H\mathbb{Z}]\cong \mathbb{Z}\langle h \rangle$? why aren't we dualise anything here as done above? (I know that homology is not necessarily the dual of cohomology, but I want to understand why we dualise above and not here, since map of spectra indices stable Cohomology operations.

$\begingroup$Homology operations and cooperations are different beasts. I'll try to write something about their relations when I have more time. Regarding (3), that is basically the definition of $H\mathbb{Z}$ and $h$ (recall: $[S^0,H\mathbb{Z}] = \pi_0H\mathbb{Z}=\mathbb{Z}$ and $h$ is just the name we gave to the generator!)$\endgroup$
– Denis NardinNov 19 '16 at 23:58

1 Answer
1

It is well known and classical that cohomology operations correspond to map of spectra: that is if $E,F$ are two spectra, a natural transformation $E^*X→F^{*+n}X$ correspond by Yoneda to a map of spectra $E→\Sigma^nF$.

On the other hand if you have a map of spectra $E→\Sigma^nF$, you also get a homology operation $E_*X→F_{*-n}X$. This is because homology can be written as
$$ E_*X = \pi_*(E\wedge X),\qquad F_{*-n}X=\pi_*(\Sigma^nF\wedge X)\,.$$
So understanding cohomology operations will give you also a supply of homology operations. This is not a dualization as you said, it is just a way to get a homology operation out of a map of spectra: no duality required. Moreover we can prove that all homology operations are of this form.

In fact any such operation is determined by its restriction to finite spectra (since every spectrum is a filtered homotopy colimit of finite spectra), and using the canonical equivalence $\mathrm{Sp}_{fin}\cong\mathrm{Sp}_{fin}^{op}$, such a natural transformation is the same thing as a natural transformation
$$E^{-*}X=E_*DX→F_{*+n}DX=F^{-*-n}X$$
when $X$ varies among finite spectra and $D$ is the Spanier-Whitehead dual. But we know that these operations are induced by maps $E→\Sigma^nF$.

Basically, when you restrict to finite spectra homology and cohomology are pretty much the same thing: you just need a Spanier-Whitehead dual to pass from one to the other.

We can say more about the second differential in the AHSS: if we look at the construction of the AHSS that uses the Postnikov tower for the spectrum $E$ instead than the one using the cellular filtration for $X$, we immediately see that the differential $d_2$ is the map of spectra given by the k-invariants of $E$. That is, we can look at the AHSS as the spectral sequence associated to the exact couple

That is the $d_2:H_*(X;\pi_nE)\to H_{*-1}(X;\pi_{n+1}E)$ is induced by the map of spectra
$$H\pi_nE\to \Sigma P_{n+1}E\to \Sigma H\pi_{n+1}E\,.$$
So we can leverage our understanding of the cohomology operations to get at the $d_2$ of the AHSS.

Now let us tackle homology cooperations. These are the homotopy classes of $E\wedge E$. I really do not have too much to say about them: they show up because they are important for the Adams spectral sequence (in fact they basically run the show there). They are also the "dual" of the cohomology operations. If $E$ is an $E_\infty$-ring spectrum we can use the universal coefficient spectral sequence to go
$$\mathrm{Ext}_{E_*}(\pi_*(E\wedge E),E_*) \Rightarrow \pi_*F_E(E\wedge E,E)=\pi_*F(E,E)$$
where $F_E$ and $F$ are the internal hom in $E$-modules and spectra respectively. For example when $E=Hk$ for $k$ a field (the main case people treat I think) this spectral sequence degenerates telling you that the $k$-linear dual of $\pi_*(Hk\wedge Hk)$ is the ring of cohomology operations. As you can see these have no direct relation to homology operations and they arise in a different setting altogether.

Finally, the Hurewicz homomorphism. This is, by definition, the map on homology theories defined by the map $h:\mathbb{S}\to H\mathbb{Z}$ sending 1 to 1. In fact, if $f:\Sigma^n\mathbb{S}\to X$ is a class in $\pi_nX$ we are sending it to the image of the fundamental homology class of $\Sigma^n\mathbb{S}$. But the fundamental homology class is just the map
$$\Sigma^nh=(h\wedge 1):\Sigma^n\mathbb{S}\to H\mathbb{Z}\wedge\Sigma^n\mathbb{S}\,.$$
So we are sending $f$ to $(1_{H\mathbb{Z}}\wedge f)\circ (h \wedge 1) = (h\wedge f)$, that is we are doing the homology operation represented by $h$.

$\begingroup$Thanks Denis, your answer is truly impressive. I will need a little bit of time to meditate on it, but I wanted to thank you as soon as possible!$\endgroup$
– RiccardoNov 20 '16 at 9:34

$\begingroup$In the first part of your answer you claim that if $X$ is finite spectrum then all the homology operations comes from a map of spectra, don't you? Because just before you said that you ignore if all of the homology operations "comes from cohomology operations" but then it seems that you prove that all homology operations are of that form.I am "stressing" this point since in the lemma I am referring to $X$ is just any spectrum, not necessarily finite$\endgroup$
– RiccardoNov 20 '16 at 9:59

2

$\begingroup$@Riccardo Sorry, there were some leftover from the first draft (when I did not realize that we could restrict to finite spectra where the problem is easy). Now it should be better. I also added an explanation of why we can restrict to finite spectra (basically: every homology theory is determined by its restriction to finite spectra, so its operations are too).$\endgroup$
– Denis NardinNov 20 '16 at 13:07